Heap's Asymmetrical String

Number Theory Level 1

Let the following number denote $H$ , where the digits are formed by concatenating all possible permutations of 10 distinct digits in increasing order (0123456789,0123456798, ...,9876543210).

$12345678901234567980123456879\dots 9876543210$

Is $H$ divisible by 11?

Yes. No.

2 solutions

Abhishek Sinha
Mar 26, 2018

Any digit $k, 0\leq k\leq 9$ appears $10!$ times in $H$ . Thus, $H=\sum_{k=0}^{9} k (\sum_{i=1}^{10!} 10^{k_i}),$ since $11| \sum_{i=1}^{10!}10^{k_i}$ , as $k$ appears equal times on the odd places and even places, it follows that $H$ is divisible by $11$ .

Giorgos K.
Mar 26, 2018

Mathematica

Mod[FromDigits@Flatten@Permutations[0~Range~9],11]

returns 0

An other way to do this is to check if the alternating sum of digits ( $1-2+3-4+5-6...3-2+1-0$ ) is divisible by $11$

For the hands-on method, you can easily calculate that by considering that each permutation consists of even digits, and that there are $9!$ of permutations with the same positioned digit (zero inclusive). The calculation turns out to be neat since the digits are subtracted by themselves.

Michael Huang - 3 years, 2 months ago

Heap's Asymmetrical String

2 solutions

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